[LeetCode] 523. Continuous Subarray Sum
grandyang opened this issue · 5 comments
Given an integer array nums and an integer k, return true ifnums has a continuous subarray of size at least two whose elements sum up to a multiple of k , orfalse otherwise.
An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.
Example 1:
Input: nums = [23,2,4,6,7], k = 6
Output: true
Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.
Example 2:
Input: nums = [23,2,6,4,7], k = 6
Output: true
Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.
Example 3:
Input: nums = [23,2,6,4,7], k = 13
Output: false
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1090 <= sum(nums[i]) <= 231 - 11 <= k <= 231 - 1
这道题给了我们一个数组和一个数字k，让求是否存在这样的一个连续的子数组，该子数组的数组之和可以整除k。遇到除法问题，肯定不能忘了除数为0的情况等处理。还有就是如何能快速的遍历所有的子数组，并且求和，这里肯定不能完全的暴力破解，OJ 肯定不答应。需要适当的优化，如果是刷题老司机的话，遇到这种求子数组或者子矩阵之和的题，应该不难想到要建立累加和数组或者累加和矩阵来做。没错，这道题也得这么做，我们要遍历所有的子数组，然后利用累加和来快速求和。在得到每个子数组之和时，先和k比较，如果相同直接返回 true，否则再判断，若k不为0，且 sum 能整除k，同样返回 true，最后遍历结束返回 false，参见代码如下（不过貌似这种方法现在已经超时了）：
解法一：
// Time Limit Exceeded (TLE)
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
for (int i = 0; i < nums.size(); ++i) {
int sum = nums[i];
for (int j = i + 1; j < nums.size(); ++j) {
sum += nums[j];
if (sum == k) return true;
if (k != 0 && sum % k == 0) return true;
}
}
return false;
}
};
下面这种方法用了些技巧，那就是，若数字a和b分别除以数字c，若得到的余数相同，那么 (a-b) 必定能够整除c。感谢热心网友 iChenLei 提供的简易证明：
a % c == b % c
a = mc + d;
b = nc + d;
a - b = mc + d - (nc + d) = (m - n) * c
so (a - b) % c == 0
明白了这条定理，用一个集合 HashSet 来保存所有出现过的余数，如果当前的累加和除以k得到的余数在 HashSet 中已经存在了，那么说明之前必定有一段子数组和可以整除k。需要注意的是k为0的情况，由于无法取余，就把当前累加和放入 HashSet 中。还有就是题目要求子数组至少需要两个数字，那么需要一个变量 pre 来记录之前的和，每次存入 HashSet 中的是 pre，而不是当前的累积和，参见代码如下：
解法二：
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
int n = nums.size(), sum = 0, pre = 0;
unordered_set<int> st;
for (int i = 0; i < n; ++i) {
sum += nums[i];
int t = (k == 0) ? sum : (sum % k);
if (st.count(t)) return true;
st.insert(pre);
pre = t;
}
return false;
}
};
既然 HashSet 可以做，一般来说用 HashMap 也可以做，这里我们建立余数和当前位置之间的映射，由于有了位置信息，就不需要 pre 变量了，之前用保存的坐标和当前位置i比较判断就可以了，参见代码如下：
解法三：
class Solution {
public:
bool checkSubarraySum(vector<int>& nums, int k) {
int n = nums.size(), sum = 0;
unordered_map<int, int> m{{0,-1}};
for (int i = 0; i < n; ++i) {
sum += nums[i];
int t = (k == 0) ? sum : (sum % k);
if (m.count(t)) {
if (i - m[t] > 1) return true;
} else m[t] = i;
}
return false;
}
};
Github 同步地址：
参考资料：
https://leetcode.com/problems/continuous-subarray-sum/
https://leetcode.com/problems/continuous-subarray-sum/discuss/99567/java-solution
https://leetcode.com/problems/continuous-subarray-sum/discuss/99499/java-on-time-ok-space
LeetCode All in One 题目讲解汇总(持续更新中...)
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这里就不证明了，博主也不会证明。明白了这条定理。
Simple proof
a % c == b % c
a = mc + d;
b = nc + d;
a - b = mc + d - (nc + d) = (m - n) * c
so (a - b) % c == 0
这里就不证明了，博主也不会证明。明白了这条定理。
Simple proof
a % c == b % c a = mc + d; b = nc + d; a - b = mc + d - (nc + d) = (m + n) * c
so (a - b) % c == 0
已添加，多谢提供～
@grandyang I am so sorry, please correct my proof.
- a - b = mc + d - (nc + d) = (m + n) * c
+ a - b = mc + d - (nc + d) = (m - n) * cAnyway, Thanks for your leetcode solutions repository, it's very helpful for me.
@grandyang I am so sorry, please correct my proof.
- a - b = mc + d - (nc + d) = (m + n) * c + a - b = mc + d - (nc + d) = (m - n) * cAnyway, Thanks for your leetcode solutions repository, it's very helpful for me.
OK, thanks!